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| author | hardythe1 | 2015-05-05 14:21:39 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-05-05 14:21:39 +0530 |
| commit | fba055ce5aa0955e22bac2413c33493b10ae6532 (patch) | |
| tree | be70ef4fccd07c9c88de778014219201b4ea971f /Materials_Science/Chapter08_1.ipynb | |
| parent | 67068710030ddd6b6c809518c34af2e04e0bf7ca (diff) | |
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diff --git a/Materials_Science/Chapter08_1.ipynb b/Materials_Science/Chapter08_1.ipynb new file mode 100755 index 00000000..cab232b8 --- /dev/null +++ b/Materials_Science/Chapter08_1.ipynb @@ -0,0 +1,179 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:12bc74eaed4218a18fa109df85a443dc199f78a3b2cafffe1198babaf1458444"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter08:Deformation of Metals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.1:pg-175"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Example 8.1: critical resolved shear stress of silver\n",
+ " \n",
+ "\n",
+ "Ts=15;#tensile stress in Mpa\n",
+ "d=[1,1,0];\n",
+ "d1=[1,1,1];\n",
+ "csda=((d[0]*d1[0])+(d[1]*d1[1])+(d[2]*d1[2]))/((math.sqrt(d[0]**2+d[1]**2+d[2]**2))*math.sqrt(d1[0]**2+d1[1]**2+d1[2]**2));#angle degree\n",
+ "d2=[0,1,1];\n",
+ "csdb=((d[0]*d2[0])+(d[1]*d2[1])+(d[2]*d2[2]))/((math.sqrt(d[0]**2+d[1]**2+d[2]**2))*math.sqrt(d2[0]**2+d2[1]**2+d2[2]**2));#angle degree\n",
+ "t=Ts*csda*csdb;#critical resolved shear stress in MPa\n",
+ "print round(t,2),\"= critical resolved shear stress in MPa\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "6.12 = critical resolved shear stress in MPa\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.2:pg-186"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Example 8.2: yield strength of material\n",
+ " \n",
+ "import numpy.linalg as lin\n",
+ "import math\n",
+ "ys1=115;# yeild strength in MN/mm**2\n",
+ "ys2=215;# yeild strength in MN/mm**2\n",
+ "d1=0.04;#diamtere in mm\n",
+ "d2=0.01;#diamtere in mm\n",
+ "A=numpy.array([[2 ,10], [1 ,10]]);\n",
+ "B=numpy.array([230,215]);\n",
+ "x=lin.solve(A,B)\n",
+ "si=x[0];# in MN/mm**2\n",
+ "k=x[1];#\n",
+ "d3=0.016;#in mm\n",
+ "sy= si +(k/math.sqrt(d3));#yeild strength for a grain size in MN/mm**2\n",
+ "print round(sy,1),\"=yeild strength for a grain size in MN/mm**2\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "173.1 =yeild strength for a grain size in MN/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.3:pg-186"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Example 8.3: yield strength of material\n",
+ "import numpy.linalg as lin\n",
+ "import math\n",
+ "ys1=120;# yeild strength in MN/mm**2\n",
+ "ys2=220;# yeild strength in MN/mm**2\n",
+ "d1=0.04;#diamtere in mm\n",
+ "d2=0.01;#diamtere in mm\n",
+ "A=numpy.array([[2 ,10], [1 ,10]]);\n",
+ "B=numpy.array([240,220]);\n",
+ "x=lin.solve(A,B)\n",
+ "si=x[0];# in MN/mm**2\n",
+ "k=x[1];#\n",
+ "d3=0.025;#in mm\n",
+ "sy= si +(k/math.sqrt(d3));#yeild strength for a grain size in MN/mm**2\n",
+ "print round(sy,1),\"= yeild strength for a grain size in MN/mm**2\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "146.5 = yeild strength for a grain size in MN/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex18.4:pg-193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 8.4 : grain diameter\n",
+ "import math \n",
+ "\n",
+ "#given data :\n",
+ "N=9; # ASTM number\n",
+ "m=8*2**N; # no. of grains [er square millimetre\n",
+ "grain=1/math.sqrt(m);\n",
+ "print round(grain,4),\"=the grain diameter(mm) \"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "0.0156 =the grain diameter(mm) \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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